0
$\begingroup$

Hope this is the right forum for this. I am reading a seminal "mathematical biophysics" paper from 1943 by MucCulloch and Pitts, and I got into a little bit of a rabbit hole trying to understand this equation in old-style logical notation. The definition of S

This is the equation in question: The equation

By following this link, I think I got most of it figured out, and I'm assuming x’) means "predecessor of x", and so the equation should be read as something like "The property S(P) holds for t if and only if P holds for Kx and t is the predecessor of x? But what does Kx mean? Doesn't this equation fulfill the definition if Kx were just replaced by x in the definition?

Thanks for any guidance on this!

$\endgroup$
3
$\begingroup$

As you can see in the References, R.Carnap (1938) is :

Rudolf Carnap, Logical Syntax of Language, Routledge (German ed.1934; 1st English ed.1937).

Language II is Carnap's versione of W&R's Principia theory of types.

The K-operators correspond to the bounded and unbounded $\mu$-operator.

The definitional axiom (it is a contextual definition) for the bounded one is (in slightly modernized symbols) :

$G(Kx_{x \le y}[F(x)]) \leftrightarrow [ (\lnot (\exists x)_{x \le y}[F(x)] \land G(0)) \lor (\exists x)_{x \le y} (F(x) \land \forall z_{z \le x} [\lnot (z=x) \to \lnot F(z)] \land G(x))].$

Thus, $(Kx)y[F(x)]$ reads : "$\text {the least (natural) number } x \text { less-equal to } y \text { such that } F$".

In the same way, $(Kx)[F(x)]$ reads: "$\text {the least (natural) number } x \text { such that } F$".


If so, the expression:

$S(P)(t) \equiv P(Kx) (t=x'),$

where $S$ is a functor that applies to properties, reads:

"$S(P)$ is a property that holds for (number) $t$ if and only if the property $P$ holds for the least number $x$ that is the predecessor of $t$ [$t=x'$ means that $t$ is the successor of $x$]".

$\endgroup$
  • $\begingroup$ Great answer, thank you so much! $\endgroup$ – Kevin Wang Jan 16 '18 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.